Please comment understood and give us a like if you got everything :)
@vishnusiddarth7953 Жыл бұрын
mass
@sudhanshushekhar4222 Жыл бұрын
Understood
@ersoumyajitpan7205 Жыл бұрын
************ more optimized ************ int findMin(vector& nums){ int left = 0, right = nums.size() - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[right]) { left = mid + 1; } else { right = mid; } } return nums[left]; }
@adil_k Жыл бұрын
East or west. Striver is the best
@CheragAli6911 ай бұрын
I came only for Minimum in Rotated Sorted Array so why I would seen previews video #striver ?
@shivajirao9996 ай бұрын
the feeling of writing the exact same code by your own without even looking at the lecture is insane, ALL thanks to your level of explanation
@ravikr34636 ай бұрын
Abe easy hai ye upper ke 3 phele se kiye hue the esiliyee btw good
@Rahul_Mongia6 ай бұрын
@@ravikr3463
@akshatgupta87872 ай бұрын
@@ravikr3463 bhai merse wo karne ke baad bhi nahi hua yrr
@hithardhksvln530221 күн бұрын
same here bro, solved one more problem in the same way.
@abhishekverma7604 Жыл бұрын
guys, this is one of the most important question if u r preparing for amazon and adobe.. hell they asked it a lot in previous 6 months according to Leetcode stats..
@animeshshaw388 Жыл бұрын
Thanks brother 🙏🏾
@akshitbhasin7322 Жыл бұрын
The fact that you acknowledged the problem of having different formats in binary search deserve a like. keep up the good work!
@HananTariq-ws9wd Жыл бұрын
Understood!! Hi I'm a Pakistani and I study at US. I really like your DSA playlist and recommend it to everyone. Thank you so much for your effort.
@AmanSingh-wr6mj10 ай бұрын
Pucha kisine?
@travelbest56234 ай бұрын
Ab ji*@d recruitment mai bhi dsa poochne lage kya? 😂😂
@vishious1411 ай бұрын
6 minutes into the video, I was able to deduce the whole logic as soon as I understood the approach !!!!!! Amazing content !!!!!!
@ArdentMusicLoverАй бұрын
Thank you, Striver. Using your approach , I was able to solve the minimum in Rotated Sorted Array-II as well , which is a LC-Hard. Hats off to you!
@surendhar.v4952 Жыл бұрын
If hurts the most , when you realize this easy question took you more than 2 hour and failed several time while in the test cases below 5 when you submit in leetcode. I did a course on java in youtube from a famous youtuber an year ago. I took a short break of two months for my semster exam and other activities. I can remind that I solved a similar type of question in an year ago. But now , I am struggling to do a easy problem like this.... Taking a break, even it is small gets your mind out of the programming mode.
@vaibhavvm4147 Жыл бұрын
bro where did u do dsa in java from?
@deepakjain4481Күн бұрын
i guess its medium
@ayushpatidar5778 Жыл бұрын
The way you cover each and every edge case is amazing. Understood everything 👍
@mehulthuletiya497 Жыл бұрын
00:38 Problem statement 01:21 Brute-force approach 01:38 Optimal approach 06:30 Dry-run 10:35 Pseudocode 12:57 Code 13:20 Optimised version : If you want try to do then 15:04 Code 15:27 Complexity
@shrirambalaji2915 Жыл бұрын
just watched like 7 min and understood the solution and what i am missing out thanks man you are awesome
@RajeshS-n2j11 күн бұрын
You are the only one who considered the case that minimum can lie in sorted half as well, that is cool. 😄Many others who provided a solution simply eliminated the sorted half without really giving any explanation.
@abhaymaurya96 ай бұрын
the same exact code i was able to write down without even watching your lecture first , kudos to u🙌❣
@iWontFakeIt Жыл бұрын
man what an explanation ! I saw several solutions on leetcode discuss but couldn't understand all of it, some if else conditions were going above my head, but you made it so clear I did code it myself both leetcode 153 and 154 with AC solutions
@AyushEditz-hs6pf3 ай бұрын
This works as well: first check if the right side of mid is sorted then check of left side. int findMin(vector& nums) { int ans=INT_MAX; int low=0; int high=nums.size()-1; while(low
@puzzle-headed-catАй бұрын
Understood! These are the kinda explanations I was looking for. Thank you so much for all the hard work and sharing with us!
@dhruvdonsahu997210 күн бұрын
this one works and its less complex: int low=0; int high=nums.size()-1; int min=INT_MAX; while(low
@bhushanambhore83782 ай бұрын
I felt proud of myself of solving this question in o(logn) time without even looking this video cz obviously I've seen your previous 2 videos and of course able to solve this one 😎.. Thank you Striver.
@prajeetdubey821322 күн бұрын
Key takeaway from this problem: 1) Basically if the array is rotated either the left or right of the mid will definitely be sorted. So make use of it. 2) Once you find the sorted subarray you definitely know that its 1st element is the smallest so put it in ans. Now no need of that subarray. (if arr[low]
@anishaa3298Ай бұрын
I was able to solve this without looking at the pseudocode 😭. thank you Striver. but i still watch till the end.
@harshitgarg2820 Жыл бұрын
Striver sir plz start a series for strings to simplify strings for us just like you did with the other topics🙏
@RagaviSathiyamoorthy Жыл бұрын
I'm really blessed to watch and learn the concept form you striver and i'm solving the problems now without referring the solutions. Thanks a lot.
@Cubeone114 ай бұрын
I used to think that neetcode gives the best explanation, but after watching this playlist i changed my thoughts.
@parth243910 ай бұрын
Understood, the last optimisation was great !!
@LifeWithSnigdha5 ай бұрын
I was solving this for the first time , was struggling because everytime was giving me runtime error and guess what i lerally watched the first 10 mins only infinite times , because my dumb brain was just not convincing myself that how is this done, at 3pm night i passed this that too in 0ms time without seeing your written code 😭😭😭😭😭. It was giving me stress because i haven't been able to pass a single solution that day so i really needed anything to get accepted just to boost my morale . Thanks striver.
@SahitiDantuluri6 ай бұрын
I have gone through some solutions to this problem and able to understand but if I try to recollect and code it later I would be a little confused. But your explanation and approach is so detailed and very easy to understand that I am able to solve it on my own without any confusion (even after many days). Thanks a lot!!
@himanshusingh18994 ай бұрын
Best DSA course on you tube.
@namangarg8976 Жыл бұрын
Another approach I thought of - -> If left array is sorted, right is unsorted. Then go to right as pivot point can be current element or in right array -> If right is sorted, left is unsorted. Then go to left as pivot point can be current element or in left array -> If both are sorted then break and just compare the ans with arr[low] as complete search space is sorted. int findMin(vector& arr) { int low = 0; int high = arr.size() - 1; int ans = INT_MAX; while(low
@ompandey291111 ай бұрын
Was wondering, What if I search for target = INT_MIN and keep the code same as the previous video wont the arr[low] be my answer?
@visase2036 Жыл бұрын
Thanks Striver. Adding my thoughts for (duplicates) . If we apply the previous logic of high-- or low++ as the mid and low/high values are equal , we will end up getting the minimum element but that does not gaurentee the no of times array has been rotated . Examples: array=[1,1,2,1,1] , orignal array = [1,1,1,1,2]. The correct answer is 3 (as the element at 0th index has been moved to 3rd index [1,1,2,1,1]). But if we apply the previous logic, the answer would come as 0 as 0th index is the minimum element. To upsolve this, we can do the following : Keep reducing high, untill [high-1] > [high] (2>1). Once you attain this point, high(3rd index) will be the answer.
@floatingpoint7629 Жыл бұрын
this does not cover all the cases
@ayushmittal9666 Жыл бұрын
I think if we remove the condition of checking if(a[low]
@deepakff74982 ай бұрын
you are binary search playlist is goated bro
@sukhjattana5887 Жыл бұрын
u unfolded the mysterious binary search....thank you!!!
@shreyanshdasgupta58072 ай бұрын
00:04 Find the minimum in a rotated sorted array using binary search. 02:08 Identify the sorted half in a rotated sorted array 04:19 The left portion of the rotated sorted array contains the minimum. 06:37 Identifying and eliminating the minimum element in a rotated sorted array. 08:40 Finding the smallest element in rotated sorted arrays using binary search. 11:06 Binary search on a rotated sorted array 13:13 Optimization in identifying sorted search space 15:16 Binary search stops in a sorted search space, improving time complexity
@Manishgupta200 Жыл бұрын
Other way to solve this problem is by pivot search.. Here is the approach, if(nums[nums.size()-1] > nums[0]) return nums[0]; // find pivot int s = 0, e = nums.size()-1, mid = s + (e - s) / 2; while(s < e){ if(nums[mid] >= nums[0]){ s = mid + 1; } else{ e = mid; } mid = s + (e - s) / 2; } return nums[mid];
@Rohitkumar-bx8ne7 ай бұрын
solved without any help with BS, but before solving i have watched the previous two videos of the playlist which helped in developing the thinking skill for solving this problem
@luvdhamija5157 Жыл бұрын
We know that if it is sorted in nature then arr[low] would always be the minimum number of all. by considering this we can simplify this as following. low=0 high=len(nums)-1 while lownums[high]: low=mid+1 elif nums[low]>nums[mid]: high=mid else: break return nums[low]
@kurma.4932Ай бұрын
Heres a alternate and easier code Got the intuition myself thankyou Striver :3 class Solution { public: int findMin(vector& nums) { int n = nums.size(); int low = 0; int high = n - 1; while ( low
@MansiBansalc9 ай бұрын
YOU ARE AN AMAZING TEACHER! THANKYOU FOR EXISTING!
@cinime Жыл бұрын
Understood! Awesome explanation as always, thank you soooo much for your effort!!
@andycharlie3255 Жыл бұрын
woow bro, this is crazy, legendary explanation, hats off man
@Manasidas99 Жыл бұрын
Understood sir thank you very much, sir. Your teaching style is really amazing. I hope I will crack my interview.
@keerthireddy8074 Жыл бұрын
I think this below code is simpler and easier to understand: int findPivot(vector &nums){ int low = 0; int high = nums.size()-1; int pivot =0; // if not rotated, it should return 0 while (low
@Aks-4711 ай бұрын
slightly lengthier, but core logic is , we need to handle 2 cases where we are uncertain where to move, that was the crux for me, a) mid is less than lo and hi, b) mid greater than lo and hi, if we exclude that , then searching is regular BS, attaching java code for the same int ans = Integer.MAX_VALUE; while(lo
@samlinus836 Жыл бұрын
I guess this algorithm fails for the below input: arr = [3,1,2] Your explanation is so clear bro❤ This condition should be added: min = Math.min(min,nums[mid]);
@theanimerecaphub Жыл бұрын
ya you have to save mid as a probable answer and compare it while moving on left
@NitinKumar-wm2dg Жыл бұрын
change it to nums[low] , you might get your answer, striver's code passes all the testcases
@anuragparasharsarmah10459 ай бұрын
How I thought about this solution is, the minimum is always in the left part of the array in a normal sorted array. However in a rotated sorted array the minimum is always in the unsorted half. So we move to the unsorted right half if it exists. Otherwise we always move to the left. while(low
@hariomtiwari9283 Жыл бұрын
Super Explanation 🎉🎉
@jagdishkhetre4515 Жыл бұрын
Understood...Awesome Binary search Playlist.. 👏
@UserUser-tn8tv11 ай бұрын
Understood. Very Good Explanation
@nirajaya55 ай бұрын
Great explanation! Thank you so much.
@paragroy5359 Жыл бұрын
Thanks a lot for making such videos. Your content is amazing. Keep on doing the great work
@sunnykumarpal5087 Жыл бұрын
Bhaiya your explanation is relay fabulous. It makes hard concepts easily understandable to us. Thank you bhaiya for helping us in dsa.
@baibhavghimire382710 ай бұрын
You beat neetcode on this🎉. Superb dude.
@Anshydv3 Жыл бұрын
understood ! the best explanation bhaiya , you are hero and a real gem❤
@aps21295 ай бұрын
Amazing explanation!!!!!
@dayashankarlakhotia4943 Жыл бұрын
Good explanation. Explanation in depth
@rahulsidhu5945 Жыл бұрын
Understood.. Awesome Binary search Playlist..😍
@tanishkthakur99657 ай бұрын
got it , understood everything
@akkipinky91949 ай бұрын
Thanks for the incredible knowledge u give us...understood!!
@AnmolGupta-oj4lm Жыл бұрын
Understood Very Well!
@per.seus._ Жыл бұрын
UNDERSTOOD❤
@myproject6768 Жыл бұрын
Absolutely understand ❤
@RishabhKumar00945 ай бұрын
understood, for duplicate elements worst case time complexity will be O(n/2).
@harishms53304 ай бұрын
how
@RishabhKumar00944 ай бұрын
@@harishms5330 So consider an example arr=11111111111111 size=14 it will take about O(7), if arr[low]==arr[mid]==arr[high] low++; high--; till low crosses high, it will follow the condition , at one time, low increases to 1 and high decreases to 1, so arr size reduced by 2 similarly it will go till O(n/2).
@harishms53304 ай бұрын
@@RishabhKumar0094 thank u🫡
@RolexGTA Жыл бұрын
i am learn from you , one day i will cross you 🙇
@kiranmoura2974 Жыл бұрын
Bahut ache se smgh aaya sir ❤
@aishezsingh70045 ай бұрын
I think This is more intutive moving towards unsorted and if sorted move left while(low arr[mid] ) high = mid - 1; else if( arr[mid] > arr[high] ) low = mid + 1; else { ans = min(ans , arr[low] ); break; } } return ans;
@dipingrover19707 ай бұрын
amazing explanation . thanks a lot
@rushidesai28367 ай бұрын
Great question!
@NazeerBashaShaik8 ай бұрын
Understood, thank you.
@md_seraj786_9 ай бұрын
understood all clear ❤
@riteshsrivastava3447 Жыл бұрын
Solution based on applying binary search on sorted side and storing start index element as possible solution : tops 100% in leetcode int findMin(vector& nums) { int ans=INT_MAX; int start=0; int end=nums.size()-1; // logic is to find sorted side and store start index value into min each time while(start
@yossihadad8558 Жыл бұрын
amazing!
@debapriyachandra1767 Жыл бұрын
Can also be done by finding pivot. class Solution { public: int findMin(vector& nums) { int n=nums.size(); int left=0,right=n-1; while(right >= left){ int mid=(left+right)>>1; int val=nums[mid]; int prev=1e9; int next=1e9; if(mid-1>=0)prev=nums[mid-1]; if(mid+1 val && next > val)return val; if(val >= nums[left] && val >= nums[right]){ left=mid+1; } else right=mid-1; } return -1; } };
@piyushroy32786 ай бұрын
understood sir. Huge kudos to you :)
@atharva_g_vlogs3 ай бұрын
UNDERSTOOD
@GodeDev11 ай бұрын
Thank you
@Shunya_Advait Жыл бұрын
Understood Sir. Thank You Sir 👌👌
@adarshkumarrao3478 Жыл бұрын
UNDERSTOOD
@samuelfrank1369 Жыл бұрын
Understood, Thanks a lot
@neerajkumar-ts6om2 ай бұрын
You don't know how relived I am to know that you won't change format, I was afraid that it is some advance technique. I always hated binary search, hope I wouldn't after this your Binary Search Playlist
@make4u9822 ай бұрын
understood good job bro
@aliakbaransaria3-925 Жыл бұрын
Very good explanation Thank you
@JothiprakashThangaraj5 ай бұрын
understood, thanks a lot!
@ddevarapaga51345 ай бұрын
Understood perfect bro
@kingbadshah45210 ай бұрын
thanks striver understood everything
@kiranmoura2974 Жыл бұрын
For duplicates I think the whole code will remain same only the optimized part will be changed as if (a[low]==a[mid] and a[low]==a[high]) {ans=min(ans,arr[low]); Low++; High--; Continue; Please check whether it is correct or not .
@anshuman-sl9du Жыл бұрын
yeah its correct dude !! thanx i was stuck there 🤖
@omkarshendge54385 ай бұрын
yea you are right so basically the duplicates case arises when there is like mid low high all three are same so basically this case is same as search in a rotated sorted array 2 right? there also this case has rose.
@RagaviSathiyamoorthy Жыл бұрын
Sir completed the problem which you gave as homework 😇😇
@bilal_khan851410 ай бұрын
Understood Bhaiya 😃
@infernogamer52 Жыл бұрын
Understood Bhaiya!
@selviraghavi-jp4ep8 ай бұрын
Minimum in rotated sorted array with duplicates my solution : class Solution { public int findMin(int[] arr) { int low = 0; int high=arr.length - 1; int min = Integer.MAX_VALUE; while(low
@saswatrath46468 ай бұрын
it's correct
@saswatrath46468 ай бұрын
thanks for sharing I was looking to match code with someone
@selviraghavi-jp4ep8 ай бұрын
Welcome bro :)@@saswatrath4646
@samratsamrat84772 ай бұрын
I first like💓💓 it then watch it because I know it is the best.😎😎
@saiyyamagrawal7653 ай бұрын
int findMin(vector& nums) { int n = nums.size(); int low = 0 , high=n-1; while(low
@venkynani9892 Жыл бұрын
My thought process: int low=0; int high=arr.length-1; int min=arr[high]; while(lowmin) { low=mid+1; } else{ min=arr[mid]; high=mid-1; } } return min;
@secondarypemail7181 Жыл бұрын
To find minimum element in rotated sorted aray with duplicates,just add the condition used before while finding an element in rotated sorted array with duplicates and the solution works fine.
@aniketbhura6131 Жыл бұрын
can you share the code
@heyOrca27115 ай бұрын
Understood! Sir
@sayak4636 Жыл бұрын
Hometask(LC-154):- class Solution { public: int findMin(vector& nums) { int s=0; int e=nums.size()-1;int ans=INT_MAX; while(s
@floatingpoint7629 Жыл бұрын
if interviewers ask you to modify it so it returns the index, this won't work
@6mahine_mein_google Жыл бұрын
@@floatingpoint7629 In duplicates we cant return index because it would have no meaning consider [3,3,3,3,3] this as example, minimum here is 3 but index is not defined because index can be anything here. Please correct me if i am wrong
@floatingpoint7629 Жыл бұрын
@@6mahine_mein_google if all are duplicates then array is not a rotated one. it is guaranteed that array is at least rotated once
@musabkhan616012 күн бұрын
Beats 100% with Log (n) time complexity. Simple 4 liner if-else statement class Solution { public int findMin(int[] nums) { int l = 0; int r = nums.length-1; int min = Integer.MAX_VALUE; while(lnums[mid]){ min = nums[mid]; } if(nums[mid]
@presidenttalks810 ай бұрын
Understood👍 thank you sir
@mahadishakkhor1239 ай бұрын
Hey god bless you bro❤
@humanity7880 Жыл бұрын
understood!
@AmanKumarSharma-de7ft Жыл бұрын
int findMin(vector& nums) { int start=0,end=nums.size()-1; while(startnums[end]?start=mid+1:end=mid;} return nums[start];} };